README_EN.md

2414. Length of the Longest Alphabetical Continuous Substring

中文文档

Description

An alphabetical continuous string is a string consisting of consecutive letters in the alphabet. In other words, it is any substring of the string "abcdefghijklmnopqrstuvwxyz".

• For example, "abc" is an alphabetical continuous string, while "acb" and "za" are not.

Given a string s consisting of lowercase letters only, return the length of the longest alphabetical continuous substring.

 

Example 1:

Input: s = "abacaba"
Output: 2
Explanation: There are 4 distinct continuous substrings: "a", "b", "c" and "ab".
"ab" is the longest continuous substring.

Example 2:

Input: s = "abcde"
Output: 5
Explanation: "abcde" is the longest continuous substring.

 

Constraints:

• 1 <= s.length <= 105
• s consists of only English lowercase letters.

Solutions

Solution 1: Two Pointers

We use two pointers $i$ and $j$ to point to the start and end of the current consecutive substring respectively. Traverse the string $s$, if the current character $s[j]$ is greater than $s[j-1]$, then move $j$ one step to the right, otherwise update $i$ to $j$, and update the length of the longest consecutive substring.

The time complexity is $O(n)$, where $n$ is the length of the string $s$. The space complexity is $O(1)$.

class Solution:
    def longestContinuousSubstring(self, s: str) -> int:
        ans = 0
        i, j = 0, 1
        while j < len(s):
            ans = max(ans, j - i)
            if ord(s[j]) - ord(s[j - 1]) != 1:
                i = j
            j += 1
        ans = max(ans, j - i)
        return ans

class Solution {
    public int longestContinuousSubstring(String s) {
        int ans = 0;
        int i = 0, j = 1;
        for (; j < s.length(); ++j) {
            ans = Math.max(ans, j - i);
            if (s.charAt(j) - s.charAt(j - 1) != 1) {
                i = j;
            }
        }
        ans = Math.max(ans, j - i);
        return ans;
    }
}

class Solution {
public:
    int longestContinuousSubstring(string s) {
        int ans = 0;
        int i = 0, j = 1;
        for (; j < s.size(); ++j) {
            ans = max(ans, j - i);
            if (s[j] - s[j - 1] != 1) {
                i = j;
            }
        }
        ans = max(ans, j - i);
        return ans;
    }
};

func longestContinuousSubstring(s string) int {
	ans := 0
	i, j := 0, 1
	for ; j < len(s); j++ {
		ans = max(ans, j-i)
		if s[j]-s[j-1] != 1 {
			i = j
		}
	}
	ans = max(ans, j-i)
	return ans
}

function longestContinuousSubstring(s: string): number {
    const n = s.length;
    let res = 1;
    let i = 0;
    for (let j = 1; j < n; j++) {
        if (s[j].charCodeAt(0) - s[j - 1].charCodeAt(0) !== 1) {
            res = Math.max(res, j - i);
            i = j;
        }
    }
    return Math.max(res, n - i);
}

impl Solution {
    pub fn longest_continuous_substring(s: String) -> i32 {
        let s = s.as_bytes();
        let n = s.len();
        let mut res = 1;
        let mut i = 0;
        for j in 1..n {
            if s[j] - s[j - 1] != 1 {
                res = res.max(j - i);
                i = j;
            }
        }
        res.max(n - i) as i32
    }
}

#define max(a, b) (((a) > (b)) ? (a) : (b))

int longestContinuousSubstring(char* s) {
    int n = strlen(s);
    int i = 0;
    int res = 1;
    for (int j = 1; j < n; j++) {
        if (s[j] - s[j - 1] != 1) {
            res = max(res, j - i);
            i = j;
        }
    }
    return max(res, n - i);
}